Question

The molecule A changes into its isomeric form B by following a first order kinetics at a temperature of 1000 K. If the energy barrier with respect to reactant energy for such isomeric transformation is 191.48 kJ mol\(^{-1}\) and the frequency factor is \(10^{20}\), the time required for 50% molecules of A to become B is ________________________ picoseconds (nearest integer). [R = 8.314 J K\(^{-1}\) mol\(^{-1}\)]

Hint:

To solve first-order rate constant problems, use the Arrhenius equation to calculate the rate constant \(K\) and then apply it to find the half-life.

Answer 1

Correct Answer: 69

For a first-order reaction, the time required for half-life is given by the equation: \[ t_{1/2} = \frac{0.693}{K} \] where \(K\) is the rate constant, which can be written as: \[ K = Ae^{-\frac{E_a}{RT}} \] Substituting the given values: \[ K = 10^{20} \times e^{-\frac{191.48 \times 10^3}{8.314 \times 1000}} = 10^{20} \times e^{-23.031} = 10^{20} \times e^{-\ln(10^{10})} = 10^{20} \times 10^{-10} = 10^{10} \, \text{sec}^{-1} \] Thus, the half-life becomes: \[ t_{1/2} = \frac{0.693}{10^{10}} = 6.93 \times 10^{-11} \, \text{sec} = 69.3 \times 10^{-12} \, \text{sec} \] \[ \Rightarrow t_{1/2} = 69 \, \text{picoseconds} \]